I can find the equation of a line using the slope-intercept form if I have two points on the line.

However I was trying to do the same with the standard form of the equation of a line, $ax+by=c$, and I can't make it happen.

The way it looks, there are three unknowns and only two equations in our system, where we would have $$ax_1+by_1=c$$$$ax_2+by_2=c$$ I'm not sure what $c$ is in this context or how to get a value for it with just two points as total information.

If it turns out it's actually not possible to find the equation of a line using the standard form, why is that the case?

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    $\begingroup$ Divide the whole equation by $b$. $\frac ab$ and $\frac cb$ are just as arbitrary as $a$ and $c$. Voila, you have only unknowns. $\endgroup$ – Don Thousand Oct 10 '18 at 1:53
  • $\begingroup$ @RushabhMehta, Thank you that works, but I still don't understand what $c$ represents geometrically I guess. $\endgroup$ – jeremy radcliff Oct 10 '18 at 2:20
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    $\begingroup$ $\frac cb$ represents the $y$ intercept $\endgroup$ – Don Thousand Oct 10 '18 at 2:23
  • $\begingroup$ @RushabhMehta, wouldn't it be $\frac{c}{b}$? Or is it the same? $\endgroup$ – jeremy radcliff Oct 10 '18 at 2:25
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    $\begingroup$ You are correct. That's a goof on my part. $\endgroup$ – Don Thousand Oct 10 '18 at 2:25

As you’ve noted, the system of equations in the unknown coefficients $a$, $b$ and $c$ is underdetermined. This means that there’s an infinite number of solutions to the system. (We know that the system has a solution since there is a line through the two points.) This shouldn’t really come as a surprise: if $ax+by+c=0$ is an equation of the line, you can multiply it by any nonzero scalar and get another equation of the same line. So, any of the solutions to the system of equations will give you a valid equation for the line. If you want to pick one in particular, a convenient choice is to choose $a$ and $b$ so that $a^2+b^2=1$ and $a\ge0$. With this choice, the normal vector $(a,b)$ is a unit vector and $|c|$ is then the distance of the line from the origin.

  • $\begingroup$ Thanks, this is the kind of explanation I was looking for and it makes sense. I changed the accepted answer to yours since it answers my question better. $\endgroup$ – jeremy radcliff Oct 10 '18 at 9:54

If you have two points start with $$y=ax+b$$

where you only have to find $a$ and $b$.

At this point you either plug your points in and get

$$ ax_1+b = y_1$$

$$ax_2+b = y_2$$ to solve for $a$ and $b$, and then you change your equation $$y-ax=b$$ or $$ax-y = -b$$ which is in standard form.

The other way is to find the slope intercept form $ y=mx+b$ and change it to the standard form by moving $mx$ to the left side of you equation.

  • $\begingroup$ Thank you, but I already know how to solve in slope-intercept form. I was looking to solve directly in standard form $ax+by=c$ without turning it into s-i form. What Rushabh Metha mentioned about dividing by $c$ makes sense, but I still don't really understand where $c$ comes from and what it really means geometrically to divide by $c$ $\endgroup$ – jeremy radcliff Oct 10 '18 at 2:19
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    $\begingroup$ With $c$ you have one degree of freedom. you can pick $ c=1$ and solve $ax+by=1$ with the given two points to find $a$ and $b$ from there. $\endgroup$ – Mohammad Riazi-Kermani Oct 10 '18 at 2:33
  • $\begingroup$ What if the line is vertical? $\endgroup$ – amd Oct 10 '18 at 4:57
  • $\begingroup$ Then the form $x=C$ is the standard form. $\endgroup$ – Mohammad Riazi-Kermani Oct 10 '18 at 5:31
  • $\begingroup$ My point is that you can’t get to that from $y=ax+b$. $\endgroup$ – amd Oct 10 '18 at 17:56

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